Prove that any finite cyclic group with more than two elements has at least two different generators.

Attempt: Let $G$ be a finite cyclic group with 3 elements, say $e,g,g^{-1}\in G$ such that $e=$identity element. If $G$ is a cyclic group then it must have a generator, say $g$. If $g$ generates $G$ then so does $g^{-1}$ since $o(g)=n$ implies $g^n=e$, and $g^{-n}=(g^{-1})^n=e.$ Therefore, $G$ has at least two generators; $g$ and $g^{-1}.$

(I know this is a possible duplicate but I am asking a question specifically about my proof. Is it correct?)

  • $\begingroup$ I think that you should say that $G$ is a finite cyclic group with at least three elements, then label three of them as $e$, $g$, and $g^-1$. The rest of the proof looks fine to me. $\endgroup$ – Andrew Stelzer Nov 5 '16 at 18:09
  • $\begingroup$ Or you could say, for n>2, there are two prime numbers less than n. $\endgroup$ – jnyan Nov 5 '16 at 18:13
  • $\begingroup$ @jnyan I cannot see the relevance of prime numbers less than $n$, and it in any case it is not true for $n=3$. $\endgroup$ – Derek Holt Nov 5 '16 at 18:57
  • $\begingroup$ Cyclic group is isomorphic to Zn. So one generator is fixed. For given Zn, prime numbers will be generator. For this case consider 1 as prime, since it will also be generator. $\endgroup$ – jnyan Nov 5 '16 at 19:15
  • 2
    $\begingroup$ The title is a duplicate, but what you are asking about your proof is OK. However, a much better idea would be to see in general that a cyclic group of order $n$ has $\phi(n)$ different generators. And $\phi(n)\ge 2$ for $n\ge 2$. $\endgroup$ – Dietrich Burde Nov 5 '16 at 19:27

As mentioned in the comments $g^{-n} = e$ does not imply that $g^{-1}$ has order $n$, only that the order of $g^{-1}$ is a divisor of $n$. You should base your proof that the number of elements of order $n$ is $\Phi(n)$ (the Euler totient function).


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